The equation $$x^2y''+pxy'+qy=0$$ where $p$ and $q$ are constants. Show that the change of independent variable given by $x=e^z$ transforms it into an equation with constant coefficients, and apply this technique to find the general solution to the following equation:
$$x^2y''+3xy'+10y=0$$
I know that we are supposed to find the second derivative of $x=e^z$ but I do not really understand how to do it. Some help would really be appreciated.
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$\begingroup$$$x=e^z\to dx=e^zdz\\\frac{dy}{dx}=\frac{dy}{e^zdz}=e^{-z}\frac{dy}{dz}\\$$for simplicity of representation,take $\frac{dy}{dx}=D_y(x)$ so $$xDy(x)=(e^z)e^{-z}D_y(z)=D_y(z)$$and $$x^2D^2_y(x)=x^2\frac{d}{dx}(\frac{dy}{dx})=\\x^2.\frac{dz}{dx}.\frac{d}{dz}(\frac{dy}{dx})=\\ x^2.\frac{dz}{dx}.\frac{d}{dz}(\frac{dy}{dz}.\frac{dz}{dx})=\\x^2(\frac{1}{x})\frac{d}{dz}(e^{-z}\frac{dy}{dz})=\\ x^2(\frac{1}{x})(-e^{-z}\frac{dy}{dz}+e^{-z}\frac{d^2y}{dz^2})=\\ xe^{-z}(\frac{d^2y}{dz^2}-\frac{dy}{dz})=\\e^z.e^{-z}(D^2_y(z)-D_y(z))=\\D^2_y(z)-D_y(z)=D(D-1)y(z)$$now put them down into equation $$x^2y''+pxy'+qy=0\\D(D-1)y(z)+pDy(z)+qy(z)=0\\(D(D-1)+pD+q)\color{red}{y(z)}=0$$ $$(D(D-1)+3D+10)y(z)=0\\r^2+2r+10=0\\r=-1\pm 3i\\y=c_1e^{r_1z}+c_2e^{r_2z}=e^{-z}(a\sin (3z)+b\cos (3z))$$
$\endgroup$ 4 $\begingroup$First, look at what happens if you apply the chain rule to look at $y'$ in terms of $z$:
$$\begin{eqnarray}y' & = & \frac{dy}{dx} \\ & = & \frac{dy}{dz} \cdot \frac{dz}{dx} \\ & = & \dot{y} \cdot \frac{1}{x} \\ & = & \frac{\dot{y}}{x}\end{eqnarray}$$
since $x = e^z$, $z = \ln x$ and hence $\frac{dz}{dx} = \frac{1}{x}$ (and using $\dot{y}$ to denote $\frac{dy}{dz}$).
If you then do something similar for $y''$, then it starts off like this:
$$\begin{eqnarray}y'' & = & \frac{d}{dx}\left(\frac{dy}{dx}\right)\\ & = & \frac{d}{dx}\left( \dot{y} \cdot \frac{1}{x} \right)\end{eqnarray}$$
and I won't run through the whole thing, but you have to use the product rule, along with the chain rule (since $\frac{d}{dx}\frac{dy}{dz} = \frac{d}{dz}\frac{dy}{dz}\cdot\frac{dz}{dx})$. Then, you take those expressions for $y'$ and $y''$ and substitute them into the DE, and you'll find that lots of stuff cancels out, and you'll get something like $\ddot{y} + (p - 1)\dot{y} + qy = 0$, which is a 2nd order linear DE with constant coefficients as required (but I'm not going to guarantee that I got those coefficients right).
Having done that, then you can take the given DE, make the substitution, solve it for a function of $z$, and from that you can get what it should be as a function of $x$.
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